The complete combustion reaction for C16H34 would be as follows:
C16H34 + 17/2O2 ------> 16CO2 + 17H2O
We are given the amount of oxygen to be used for the reaction. This value would be the stating point for the calculations.
0.50 mol O2 ( 1 mol C16H34 / 17/2 mol O2 ) ( 22.4 L / 1 mol ) = 1.317 L = 1317 mL C16H34
The reaction that should be followed is
Na2SO4 + C<span>a(NO3)2 --> CaSO4 + 2NaNO3</span>
first calculate the limiting reactant
mol Na2SO4 = 0.075 L (<span>1.54×10−2 mol / L) = 1.155x10-3 mol
mol Ca(NO3)2 = 0.075 L (</span><span>1.22×10−2 mol / L) = 9.15x10-4 mol
so the limiting reactant is the Ca(NO3)2
so all of the Ca2+ will be precipitated, percentage unprecipitated = 0.00 % </span>
Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
PbCl2 and PbCl4
Empirical formula is when the molecular formula is reduced (each element) by the same number until you are unable to do so any further.
CH2O cannot be further reduced, but C3H6O3 can all be divided by 3 making it CH2O (same empirical formula)
Since the heat of combustion of butane is higher than that of propane, butane will produce more heat per gram when burned compared to propane.
The heat of combustion of a compound refers to the heat evolved when one mole of the compound is burnt under standard conditions. This heat of combustion increases from methane upwards.
Since the heat of combustion of alkanes increases according to increasing molar mass, butane will produce more heat per gram when burned compared to propane.
Learn more about heat of combustion: brainly.com/question/25312146